Calculate forces on a fixed beam

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A fixed beam ABC is subjected to an evenly distributed load q and a diagonal load F at point C, requiring the calculation of forces Ax, Ay, and the reaction couple MA. The discussion highlights the importance of correctly summing forces and moments, emphasizing that moment arm distances should be considered when calculating the effects of the loads. Participants note that the resultant load from the distributed force q should be calculated as q multiplied by the length over which it acts, and it acts at the center of gravity of that load. Additionally, the correct approach for calculating the moment due to the diagonal load F involves using the sine of the angle and the appropriate distance. Accurate sign usage in calculations is crucial to avoid errors in the final results.
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Homework Statement



A beam ABC is fixed in point A. There is an evenly distributed load q working on top of the beam. In addition there is a diagonal load F working at point C.

Calculate force Ax, Ay and the reaction couple MA.




The Attempt at a Solution



Please see attached file for figures and calculations.
I believe I'm on the right path, but I fear there might be operator mistakes(+ and -) somewhere.
(Ironically that's what I find to be the most difficult thing with statics. Get a + wrong and all your calculations are off...)
 

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dreamliner said:

Homework Statement



A beam ABC is fixed in point A. There is an evenly distributed load q working on top of the beam. In addition there is a diagonal load F working at point C.

Calculate force Ax, Ay and the reaction couple MA.




The Attempt at a Solution



Please see attached file for figures and calculations.
I believe I'm on the right path, but I fear there might be operator mistakes(+ and -) somewhere.
(Ironically that's what I find to be the most difficult thing with statics. Get a + wrong and all your calculations are off...)
Oh those pesty plus and minus signs are always cause for concern in Physics and Engineering, but that is not your problem here in your incorrect solution. You are mixing up forces and moment calculations. When you sum forces in the y direction, there are no moment arm distances to consider. Check your value for the resultant force of the distributed load. And when you sum moments, check your value for the moment about A of the force F applied at C.
 
PhanthomJay said:
When you sum forces in the y direction, there are no moment arm distances to consider.

So in this case it should be only q*4,1(the length of the beam force q is working on) even though force q isn't distributed across the entire beam? (I was under the assuption you had to consider cg of force q in such cases)

PhanthomJay said:
And when you sum moments, check your value for the moment about A of the force F applied at C.

Should have been a moment arm distance there, yes. So the correct value would be F*sin 54,4*6,3
 
dreamliner said:
So in this case it should be only q*4,1(the length of the beam force q is working on) even though force q isn't distributed across the entire beam?
Correct. The load distribution is in units of force per unit length, and thus the resultant load, which must be in force units, is ql, where l is the length of the beam over which q is applied.
(I was under the assuption you had to consider cg of force q in such cases)
Once you calculate the resultant load of the force distribution, ql, the resultant acts at the cg of the load, and the magnitude and location of that resultant is used to determine end reactions when summing moments about any point.
Should have been a moment arm distance there, yes. So the correct value would be F*sin 54,4*6,3
yes.
 
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